Question: Divide the polynomials. The form of your answer should either be $p(x)$ or $p(x)+\dfrac{k}{x-3}$ where $p(x)$ is a polynomial and $k$ is an integer. $\dfrac{x^3-4x-15}{x-3}=$
Solution: Usually, there are many different ways to divide polynomials. Here, we will use the method of polynomial long division. Notice that the expression in the numerator is missing a $2^{\text{nd}}$ degree term. To avoid any confusion, let's add that term as $0x^2$. $\begin{array}{r} x^2+3x+\phantom{1}5 \\ x-3|\overline{x^3+0x^2-4x-15} \\ \mathllap{-(}\underline{x^3-3x^2\phantom{-4x-15}\rlap )} \\ 3x^2-4x-15 \\ \mathllap{-(}\underline{3x^2-9x\phantom{-15}\rlap )} \\ 5x-15 \\ \mathllap{-(}\underline{5x-15\rlap )} \\ 0 \end{array}$ We found that the quotient is $x^2+3x+5$ and the remainder is $0$, which means the answer is simply a polynomial (no expression of the form $\dfrac{k}{x-3}$ ). $\dfrac{x^3-4x-15}{x-3}=x^2+3x+5$